Water of mass 150g at 60 ^o c is added to 300g of water at 20 ^o c and the mixture is well stirred. Calculate…

Analysis

This is a classic calorimetry problem based on the Principle of Conservation of Energy (specifically, the Principle of Mixtures). The principle states that in an isolated system, the heat lost by the hot body is equal to the heat gained by the cold body.

Concept

The formula for heat energy is:

Where:

• $m$ = mass
• $c$ = specific heat capacity of water (which cancels out since both substances are water)
• $\Delta\theta$ = change in temperature

Let $T$ be the final temperature of the mixture.

• For the hot water: Heat lost = $m_1c(T_1 - T)$
• For the cold water: Heat gained = $m_2c(T - T_2)$

Solution

Given:

• Mass of hot water ($m_1$) = 150g
• Initial temperature of hot water ($T_1$) = 60°C
• Mass of cold water ($m_2$) = 300g
• Initial temperature of cold water ($T_2$) = 20°C

Step 1: Set up the equation Heat lost = Heat gained

Step 2: Simplify Since $c$ (specific heat capacity of water) is on both sides, it cancels out:

Step 3: Solve for T Divide both sides by 150 to make the calculation easier:

Rearrange the terms:

Rounding to the nearest whole number as provided in the options, we get 33°C.

Conclusion

The calculated temperature of the mixture is approximately 33°C, which matches option A.

Exam Tip

When mixing two quantities of the same substance, you can use the weighted average formula:

Using this: $T = \frac{(150 \times 60) + (300 \times 20)}{150 + 300} = \frac{9000 + 6000}{450} = \frac{15000}{450} = 33.33^\circ\text{C}$. This is a faster way to solve mixture problems in JAMB!